In Young's double slit experiment, the fringe width with light of wavelength 6000 Ã… is 3 mm. The fringe width, when the wavelength of light is changed to 4000 Ã… is

Options

A. 3 mm
B. 1 mm
C. 2 mm
D. 4 mm

Explanation

To solve the problem of finding the fringe width in Young's double slit experiment when the wavelength of light is changed, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for Fringe Width**:
The fringe width (β) in Young's double slit experiment is given by the formula:
\[ \beta = \frac{\lambda D}{d} \]
where:
- \( \beta \) = fringe width
- \( \lambda \) = wavelength of light
- \( D \) = distance from the slits to the screen
- \( d \) = distance between the slits

2. **Identify Given Values**:
From the problem, we know:
- Wavelength \( \lambda_1 = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} \)
- Fringe width \( \beta_1 = 3 \, \text{mm} = 3 \times 10^{-3} \, \text{m} \)
- New wavelength \( \lambda_2 = 4000 \, \text{Å} = 4000 \times 10^{-10} \, \text{m} \)

3. **Establish the Relationship**:
Since \( D \) and \( d \) remain constant, we can say that the fringe width is directly proportional to the wavelength:
\[ \frac{\beta_2}{\beta_1} = \frac{\lambda_2}{\lambda_1} \]

4. **Rearranging the Equation**:
From the above relationship, we can express the new fringe width \( \beta_2 \) as:
\[ \beta_2 = \beta_1 \cdot \frac{\lambda_2}{\lambda_1} \]

5. **Substituting the Values**:
Now, substitute the known values into the equation:
\[ \beta_2 = 3 \times 10^{-3} \cdot \frac{4000 \times 10^{-10}}{6000 \times 10^{-10}} \]

6. **Simplifying the Expression**:
Simplifying the fraction:
\[ \beta_2 = 3 \times 10^{-3} \cdot \frac{4000}{6000} = 3 \times 10^{-3} \cdot \frac{2}{3} = 2 \times 10^{-3} \, \text{m} \]

7. **Converting to mm**:
Convert \( \beta_2 \) back to mm:
\[ \beta_2 = 2 \, \text{mm} \]

### Final Answer:
The fringe width when the wavelength of light is changed to 4000 Å is **2 mm**.

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